Question

lim x>0 (cube root(1+2x) -1)/x

Answer 1

\[\frac{ \sqrt[3]{1+2x} -1}{ x }\]

Answer 2

do you know l'hospital ?

Answer 3

yes, please if possible it's not in our module so the answer wont be acceptable

Answer 4

make it into a difference of cubes

Answer 5

multiply on top and below the right factors to give the formula
(a-b)(a^2+ab+b^2)
in your case\[a=\sqrt[3]{1+2x},~~b=1\]

Answer 6

thanx @TuringTest . let me see

Answer 7

am stuck here @TuringTest , whi ch ones are the right factors

Answer 8

if\[a=\sqrt[3]{1+2x},~~b=1\]then\[a^2+ab+b^2=(1+2x)^{2/3}+(1+2x)^{1/3}+1\]multiply that above and below, and usethe identity \[a^3-b^3=(a-b)(a^2+ab+b^2)\]

Answer 9

i have done that, question is,now, what is next? what should i do?

Answer 10

what are you left with?

Answer 11

Show what you have so far please, so I can help you. If you did what I told you the cancellations should be obvious and you can just take the limit.

Answer 12

i am sending a pic

Answer 13

hope you can see it @TuringTest

Answer 14

that's correct, now use the identity \[a^3-b^3=(a-b)(a^2+ab+b^2)\]on the top

Answer 15

this is just an alternative, hope can help you too
let u = 1+2x -----> x = (u-1)/2
if x->0 then u -> 1

so, the limit becomes
lim u->1 (u^1/3 - 1)/((u-1)/2)
= lim u->1 2*(u^1/3 - 1)/(u-1)
factor out the denominator, get
= lim u->1 2*(u^1/3-1)/[(u^1/3-1)(u^2/3+u^1/3+1)]

now you can cancel out the similar terms, get
= lim u->1 2/(u^2/3+u^1/3+1)

finally, just subtitute u = 1
now, get your answer!

Answer 16

thanx @RadEn - i will definatly pursue it.. I am equally interested in the method at hand. so which one becomes my a and b on the numerator @TuringTest . should i fatorise or expand? or is \[(1 +2x)^{\frac{ 1 }{ 3}} -1)\] =a and the other parenthesis b ?

Answer 17

on the numerator

Answer 18

no expanding, just run the identity as I wrote it from right to left\[{(1+2x)^{1/3}[(1+2x)^{2/3})+(1+2x)^{1/3}+1]\over x[(1+2x)^{2/3})+(1+2x)^{1/3}+1]}\]noticing that the numerator is of the form\[(a-b)(a^2+ab+b^2)\] with \(a=(1+2x)^{1/3}\) and \(b=1\) we can rewrite it in the form \(a^3-b^3\)

Answer 19

ok

Answer 20

\[\frac{(1+2x)^{\frac{ 2 }{ 3}}-1) }{ x \left[ (1+2x)^{\frac{ 2 }{ 3 }}+(1+2x)^{\frac{ 1 }{ 3 } }+1 \right] }\] yeh am not gettin it

Answer 21

how did you get that numerator?
what is a?
what is a^3
what is b?
what is b^3?

Answer 22

but this is big stuff so i want to know it

Answer 23

honestly, i just took the b you gave to me and squared it, same with the a

Answer 24

i know the difference of ubes, but i am failing to figure it out in this one

Answer 25

why square it? we are using difference of *cubes*. difference of squares looks totally different, you ought to memorize those

Answer 26

\[a^3-b^3=(a-b)(a^2+ab+b^2)\]
what is a?
what is a^3?

Answer 27

oh wait!

Answer 28

@TuringTest - u are a genius man. I have realised that the numerator will be 1 +2x-1 =2x, the x cancels with the one on the denominator . this leaves us with 2 on top, and then we plug in zero to get 1 + 1+ 1 , so 2/3 is the answer thanx!

Answer 29

welcome!
These kinds of problems often have people going straight to l'hospital, bit l'hospital is actually invalid in this case anyway. either way, be sure to try out all the little algebra tricks that seemed useless (diff square, diff cubes, complete square, etc). That's usually the ticket ;)

Answer 30

thanx a million times. have just learnt something SERIOUSLY NEW today!

Answer 31

that's better than a million medals :)

Answer 32

@RadEn